 Hi, well, I'm Professor Steven Nesheva, and I want to tell you a little bit about how to integrate the Clapeyron equation along the solid-liquid-phase boundary. So just to orient you, we have a phase diagram here, temperature on this axis, pressure on that axis, and each one of these phase boundaries is a line of equilibrium between different phases, so that would be the solid-liquid and gas regions, and there's the triple point right there, which will be kind of our anchor for these integrations, and each has a slope, which will be called dp dt, the slope in the pressure direction, and right now I'm going to focus on the solid-liquid-phase boundary. Now the Clapeyron equation itself looks like this. It just says this, if I march along in the temperature direction, a small amount dt, then the pressure will go up by a certain amount, and how much will it go up? Well, it's going to go up by this factor. That's the delta H of the transition, which in this case would be the enthalpy of fusion, which we're going to consider constant for this process, but just so you have a pictorial on that. That's the change right at the phase boundary from solid to liquid. There would also be a volume change in going from the solid to the liquid, so that would be delta V of fusion, and then of course there's the temperature. Now because we have solids and liquids in there barely incompressible, the change in volume doesn't depend very much on temperature and pressure, especially around the triple point, and as I say the enthalpy doesn't change very much either, as long as we're not changing the temperature too much. So we can just integrate this equation straight up. So dp integrated becomes p evaluated from p3, our starting point, up to what I'll call p star ice liquid. p star ice liquid is, is that line, and that of course will just integrate out to being the pressure p star ice liquid along that line, wherever, whatever temperature we want to go to, minus the starting point, which is triple point pressure. So that's the left-hand side of that equation. On the right-hand side of that equation, we have this, it's, that's, we're going to consider that a constant. So the integral of 1 over t dt is just log of t, and that, we would also integrate from t3 out to some arbitrary temperature. And because the way logs work, that's going to be the log of t minus log of t3, which turns into that ratio. So the right-hand side of the equation turns into this, and as I say we had the left-hand side of the equation looking like that. Therefore, if you want, you can say the pressure along that base boundary, which I move this to the right, would be the triple point pressure plus that correction factor. And this is the Thomson equation right here, and again the Thomson equation is what gives you the, the phase boundary, the equation for the phase boundary along a solid-liquid phase equilibrium boundary, okay?